使用Python求解数独

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之前学习xlrd和xlwt模块的时候,突发奇想,能不能用Python写个求解数独的小程序呢?然后找找资料,在实验楼上发现了现成的例子,然后花了一晚上,改动了一部分,自己写了一遍。(这里的数独并不满足对角线原则,试了下,把对角线的要求加进去,通过这种方法生成一个数独太耗时间了,你们要是有兴趣,可以改进下这里)
直接上程序吧,下次有空过来整理整理。

import randomimport itertoolsfrom copy import deepcopytest = [[None, None, 3, None, 6, 9, None, 1, 7],        [None, 1, None, 2, 4, 8, None, None, 9],        [4, None, 9, 7, 1, None, None, 2, 6],        [9, 7, 2, 1, 8, 4, 6, 3, 5],        [1, None, 8, None, 5, 2, 7, 9, 4],        [6, 4, 5, 9, 3, 7, None, None, None],        [None, 9, None, 4, None, None, 2, None, 8],        [None, 6, None, 8, None, 1, None, 4, 3],        [None, None, None, None, None, None, 5, 7, 1]]'''+++&&&&&++&&&&&++&&&&&+++&&&&&++&&&&&++&&&&&+++&&&&&++&&&&&++&&&&&+++&&&     ||     ||  3  &&&     ||  6  ||  9  &&&     ||  1  ||  7  &&&+++-----++-----++-----+++-----++-----++-----+++-----++-----++-----+++&&&     ||  1  ||     &&&  2  ||  4  ||  8  &&&     ||     ||  9  &&&+++-----++-----++-----+++-----++-----++-----+++-----++-----++-----+++&&&  4  ||     ||  9  &&&  7  ||  1  ||     &&&     ||  2  ||  6  &&&+++&&&&&++&&&&&++&&&&&+++&&&&&++&&&&&++&&&&&+++&&&&&++&&&&&++&&&&&+++&&&  9  ||  7  ||  2  &&&  1  ||  8  ||  4  &&&  6  ||  3  ||  5  &&&+++-----++-----++-----+++-----++-----++-----+++-----++-----++-----+++&&&  1  ||     ||  8  &&&     ||  5  ||  2  &&&  7  ||  9  ||  4  &&&+++-----++-----++-----+++-----++-----++-----+++-----++-----++-----+++&&&  6  ||  4  ||  5  &&&  9  ||  3  ||  7  &&&     ||     ||     &&&+++&&&&&++&&&&&++&&&&&+++&&&&&++&&&&&++&&&&&+++&&&&&++&&&&&++&&&&&+++&&&     ||  9  ||     &&&  4  ||     ||     &&&  2  ||     ||  8  &&&+++-----++-----++-----+++-----++-----++-----+++-----++-----++-----+++&&&     ||  6  ||     &&&  8  ||     ||  1  &&&     ||  4  ||  3  &&&+++-----++-----++-----+++-----++-----++-----+++-----++-----++-----+++&&&     ||     ||     &&&     ||     ||     &&&  5  ||  7  ||  1  &&&+++&&&&&++&&&&&++&&&&&+++&&&&&++&&&&&++&&&&&+++&&&&&++&&&&&++&&&&&+++'''def full_board():    board = None    while board is None:        board = attmpt_board()    return boarddef attmpt_board():    nums = list(range(1,10))    board = [[None for _ in range(9)] for _ in range(9)]    for i,j in itertools.product(range(9),repeat=2):#这个函数可以将两层甚至多层循环简写一下        i0,j0 = i - i % 3,j - j % 3        random.shuffle(nums)        for x in nums:            if(x not in board[i]            and all(x != row[j] for row in board)            and all(x not in row[j0:j0 + 3] for row in board[i0:i])):                board[i][j] = x                break        else:            return None    return boarddef get_board(full_board,level=1): #难度等级    board = deepcopy(full_board)    omit = [0,35,60,81] #挖出的方块数,因为随机挖的时候可能重复挖到某个,所以一般小于该数    for _ in range(omit[level]):        i = random.randint(0,8) #随机取0——8中的一个数        j = random.randint(0,8)        board[i][j] = None    return boarddef print_board(board):    spacer = '+++-----++-----++-----+++-----++-----++-----+++-----++-----++-----+++'    for i in range(9):        if i % 3 ==0:            print(spacer.replace('-','&'))        else:            print(spacer)        print('&&&  {}  ||  {}  ||  {}  &&&  {}  ||  {}  ||  {}  &&&  {}  ||  {}  ||  {}  &&&'            .format(*(cell or ' ' for cell in board[i])))    print(spacer.replace('-','&'))def isfull(board):    for i,j in itertools.product(range(9),repeat=2):        if board[i][j] == None:            return False    return Truedef list_cell(board,x,y): #返回某个空格可以填入的数字的列表    nums = set(range(1,10))    row = set(board[x]) #所在的行已有的数字    col = set(row[y] for row in board) #列    x0,y0 = x - x % 3,y - y % 3    block = set(board[i][j] for i,j in itertools.product(range(x0,x0+3),range(y0,y0+3))) #块    result = nums - row - col - block    return list(result)def min_pos(board): #返回最短的可填列表所在的坐标    min_,x,y = 9,-1,-1    for i,j in itertools.product(range(9),repeat=2):        if board[i][j] == None:            temp = len(list_cell(board,i,j))            if temp < min_:                min_ = temp                x,y = i,j    return x,ydef next_pos(x,y): #返回下个位置的坐标    pos = 9 * x + y    x = ((pos+1) % 81)//9    y = (pos+1) % 9    return x,ydef solve_board(board):    if isfull(board):        return board    while True:        x,y = min_pos(board)        nums = list_cell(board,x,y)        if len(nums) == 1:            board[x][y] = nums[0]        else:            break    x,y = min_pos(board)    result = core_fun(board,x,y)    return resultdef core_fun(board,x,y):    if isfull(board):        return board    while board[x][y]:        x,y = next_pos(x,y)    numbers = list_cell(board,x,y)    if len(numbers) == 0:        return False    for num in numbers:        board[x][y] = num        flag = core_fun(board,next_pos(x,y)[0],next_pos(x,y)[1])#一直往下递归        if flag == False: #递归到某个空任何数都不能填进去的时候            board[x][y] = None #因为上一步操作是让board[x][y]=n,表明赋值为n是不对的。接着往下走,取numbers中的下一个数        else:            return board #board不断的在改变,直到求解出答案    #这里不好理解,这表明numbers中的所有数据用完了,还不对,怎么可能呢?    #通过上面的numbers = list_cell(board,x,y)算出来的numbers肯定是对的啊。所以肯定到不了这一步啊    #其实吧,因为这里有递归,当前层的numbers是在上一层选了一个数字之后得出的,有可能上一层选的数字就不对    #所以这里试将上一层的那个错误的数置为空    board[x][y] = None    return FalseFullBoard = full_board()print('fullboard:\n')print_board(FullBoard)print()my_board = get_board(FullBoard,3)print('my_board:\n')print_board(my_board)result = solve_board(my_board)print('\nanswer:\n')print_board(result)

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(可见,数独的答案可能并不唯一,下次试试求解所有的满足要求的答案)

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